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Integrated math 3

Course: Integrated math 3 > Unit 4

Lesson 4: Putting it all together
Math
Integrated math 3
Polynomial graphs
Putting it all together
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Graphs of polynomials

Google Classroom
Analyze polynomials in order to sketch their graph.

What you should be familiar with before taking this lesson

The end behavior of a function f describes the behavior of its graph at the "ends" of the x-axis. Algebraically, end behavior is determined by the following two questions:
  • As x→+∞, what does f(x) approach?
  • As x→−∞, what does f(x) approach?
If this is new to you, we recommend that you check out our end behavior of polynomials article.
The zeros of a function f correspond to the x-intercepts of its graph. If f has a zero of odd multiplicity, its graph will cross the x-axis at that x value. If f has a zero of even multiplicity, its graph will touch the x-axis at that point.
If this is new to you, we recommend that you check out our zeros of polynomials article.

What you will learn in this lesson

In this lesson, we will use the above features in order to analyze and sketch graphs of polynomials. We will then use the sketch to find the polynomial's positive and negative intervals.

Analyzing polynomial functions

We will now analyze several features of the graph of the polynomial f(x)=(3x−2)(x+2)2.

Finding the y-intercept

To find the y-intercept of the graph of f, we can find f(0).
f(x)=(3x−2)(x+2)2f(0)=(3(0)−2)(0+2)2f(0)=(−2)(4)f(0)=−8
The y-intercept of the graph of y=f(x) is (0,−8).

Finding the x-intercepts

To find the x-intercepts, we can solve the equation f(x)=0.
f(x)=(3x−2)(x+2)20=(3x−2)(x+2)2
↙↘3x−2=0orx+2=0Zero product propertyx=23orx=−2
The x-intercepts of the graph of y=f(x) are (23,0) and (−2,0).
Our work also shows that 23 is a zero of multiplicity 1 and −2 is a zero of multiplicity 2. This means that the graph will cross the x-axis at (23,0) and touch the x-axis at (−2,0).

Finding the end behavior

To find the end behavior of a function, we can examine the leading term when the function is written in standard form.
Let's write the equation in standard form.
f(x)=(3x−2)(x+2)2f(x)=(3x−2)(x2+4x+4)f(x)=3x3+12x2+12x−2x2−8x−8f(x)=3x3+10x2+4x−8
The leading term of the polynomial is 3x3, and so the end behavior of function f will be the same as the end behavior of 3x3.
Since the degree is odd and the leading coefficient is positive, the end behavior will be: as x→+∞, f(x)→+∞ and as x→−∞, f(x)→−∞.

Sketching a graph

We can use what we've found above to sketch a graph of y=f(x).
Let's start with end behavior:
  • As x→+∞, f(x)→+∞.
  • As x→−∞, f(x)→−∞.
This means that in the "ends," the graph will look like the graph of y=x3.
The ends of a polynomial are graphed on an x y coordinate plane. The first end curves up from left to right from the third quadrant. It is labeled As x goes to negative infinity, f of x goes to negative infinity. The other end curves up from left to right from the first quadrant. It is labeled As x goes to positive infinity, f of x goes to positive infinity.
Now we can add what we know about the x-intercepts:
  • The graph touches the x-axis at (−2,0), since −2 is a zero of even multiplicity.
  • The graph crosses the x-axis at (23,0), since 23 is a zero of odd multiplicity.
The parts of a polynomial are graphed on an x y coordinate plane. The first end curves up from left to right from the third quadrant. The other end curves up from left to right from the first quadrant. A point is on the x-axis at (negative two, zero) and at (two over three, zero). A part of the polynomial is graphed curving up to touch (negative two, zero) before curving back down. Another part of the polynomial is graphed curving up and crossing the x-axis at the point (two over three, zero).
Finally, let's finish this process by plotting the y-intercept (0,−8) and filling in the gaps with a smooth, continuous curve.
While we don't know exactly where the turning points are, we still have a good idea of the overall shape of the function's graph!
The parts of a polynomial are graphed on an x y coordinate plane. The first end curves up from left to right from the third quadrant. The other end curves up from left to right from the first quadrant. A point is on the x-axis at (negative two, zero) and at (two over three, zero). A part of the polynomial is graphed curving up to touch (negative two, zero) before curving back down. Another part of the polynomial is graphed curving up and crossing the x-axis at the point (two over three, zero). There is a point at (zero, negative eight) labeled the y-intercept. The parts of the polynomial are connected by dashed portions of the graph, passing through the y-intercept.

Positive and negative intervals

Now that we have a sketch of f's graph, it is easy to determine the intervals for which f is positive, and those for which it is negative.
A polynomial is graphed on an x y coordinate plane. The graph curves up from left to right touching the x-axis at (negative two, zero) before curving down. It curves back up and passes through the x-axis at (two over three, zero). Where x is less than negative two, the section below the x-axis is shaded and labeled negative. Where x is greater than negative two and less than two over three, the section below the x-axis is shaded and labeled negative. Where x is greater than two over three, the section above the x-axis is shaded and labeled positive.
We see that f is positive when x>23 and negative when x<−2 or −2<x<23.

Check your understanding

1) You will now work towards a sketch of g(x)=(x+1)(x−2)(x+5) on your own.
a) What is the y-intercept of the graph of g(x)=(x+1)(x−2)(x+5)?
(0,
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
)

b) What is the end behavior of the graph of g(x)=(x+1)(x−2)(x+5)?
Choose 1 answer:

c) What are the x-intercepts of the graph of g(x)=(x+1)(x−2)(x+5)?
Choose 1 answer:

d) Which of the following graphs could be the graph of g(x)=(x+1)(x−2)(x+5)?
Choose 1 answer:

2) Which of the following could be the graph of y=(2−x)(x+1)2
Choose 1 answer:

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  • blobby green style avatar for user bdenne14
    Posted 8 years ago. Direct link to bdenne14's post “How do you match a polyno...”
    How do you match a polynomial function to a graph without being able to use a graphing calculator?
    (18 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      Posted 8 years ago. Direct link to Stefen's post “Seeing and being able to ...”
      Incredible Answer
      Great Answer
      Good Answer
      Seeing and being able to graph a polynomial is an important skill to help develop your intuition of the general behavior of polynomial function. In practice, we rarely graph them since we can tell a lot about what the graph of a polynomial function will look like just by looking at the polynomial itself.
      For example, given ax² + bx + c
      If a is positive, the graph will be like a U and have a minimum value.
      If a is negative, the graph will be flipped and have a maximum value
      If |a| is > 1 the parabola will be very narrow.
      If |a| is < 1 the parabola will be very wide.
      The axis of symmetry (and the location of the vertex) is given by -b/2a.
      When you factor the function, you get the x-intercepts (if any).
      When you set x=0 you get the y intercept of the graph (if any).
      Check out these pages for more information:
      https://www.mathsisfun.com/algebra/quadratic-equation-graphing.html

      Here is a neat tool where you can change the values of the coefficients of a polynomial and see how it affects the resulting graph - a really great way to develop your "mind's eye" and visualize the graph just by looking at the polynomial.
      https://www.mathsisfun.com/algebra/quadratic-equation-graph.html
      (55 votes)
  • male robot hal style avatar for user Reginato Rezende Moschen
    Posted 7 years ago. Direct link to Reginato Rezende Moschen's post “What is multiplicity of a...”
    What is multiplicity of a root and how do I figure out?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Alissa
      Posted 6 years ago. Direct link to Alissa's post “When you have a factor th...”
      When you have a factor that appears more than once, you can raise that factor to the number power at which it appears. For example if you have (x-4)(x+3)(x-4)(x+1)
      the multiplicities of each factor would be:
      x-4= multiplicity 2
      x+3+ multiplicity 1
      x+1= Multiplicity 1
      (8 votes)
  • duskpin seed style avatar for user bavila470
    Posted 6 years ago. Direct link to bavila470's post “Can there be any easier e...”
    Can there be any easier explanation of the end behavior please. The way that it was explained in the text, made me get a little confused. Also, for the practice problem, when ever x equals zero, does it mean that we only solve the remaining numbers that are not zeros?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user john.cueva
    Posted 3 years ago. Direct link to john.cueva's post “How can you graph f(x)=x^...”
    How can you graph f(x)=x^2 + 2x - 5? step by step?
    (I'm new here)
    (5 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user A/V
      Posted 3 years ago. Direct link to A/V's post “Given a polynomial in tha...”
      Given a polynomial in that form, the best way to graph it by hand is to use a table.

      My personal rule is to always use the values most easiest, so let's use -2,-1,0,1,2

      F(-2) = (-2)²-2(-2)-5 = 4+4-5 = 3 ==> (-2,3)
      F(-1) = (-1)²-2(-1)-5 = 1+2-5 = -2 ==> (-1,-2)
      F(0) = (0)² -2(0) -5 = 0-0-5 = -5 ==> (0,-5)
      F(1) = (1)² -2(1) -5 = 1-2-5 ==> (1,-6)
      F(2) = (2)² -2(2)-5 = 4-4-5 = (2,-5)


      ^ just a reminder that the number inside the f(x) (e.g f(-1)) is the number that you plug in the equation.

      As you notice, the value inside the f(x) is the x-coordinate, and the output of that is the y coordinate
      (e.g (-1,-2) at f(-1)).

      hopefully that helps ! and if you have any suspicious, 100% recommend using desmos to check your work.
      (4 votes)
  • piceratops ultimate style avatar for user Cătălin Gherasim Cîrcu
    Posted 6 years ago. Direct link to Cătălin Gherasim Cîrcu's post “What throws me off here i...”
    What throws me off here is the way you gentlemen graphed the Y intercept. Surely there is a reason behind it but for me it is quite unclear why the scale of the y intercept (0,-8) would be the same as (2/3,0). Would appreciate an answer. Have a good day!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • aqualine tree style avatar for user Judith Gibson
      Posted 6 years ago. Direct link to Judith Gibson's post “I see what you mean, but ...”
      I see what you mean, but keep in mind that although the scale used on the X-axis is almost always the same as the scale used on the Y-axis, they do not HAVE TO BE the same.
      I can't answer for the designers of the graph, but perhaps they wanted to save space, or perhaps it was done in error --- who knows!
      (7 votes)
  • blobby green style avatar for user manasvi
    Posted 7 months ago. Direct link to manasvi's post “How do you find the end b...”
    How do you find the end behavior of your graph by just looking at the equation. how do you determine if it is to be flipped?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • winston baby style avatar for user noedig101
      Posted 7 months ago. Direct link to noedig101's post “by looking at the highest...”
      by looking at the highest degree term.
      kx^n
      if n is even, the two "ends" will be the same. if it's odd they will be different.
      k determines what infinity is.
      if k is positive, infinity will be positive.
      if k is negative, infinity will be negative.

      combine the two to figure out negative infinity.
      (4 votes)
  • area 52 yellow style avatar for user Coward
    Posted 2 years ago. Direct link to Coward's post “Question number 2--'which...”
    Question number 2--'which of the following could be a graph for y = (2-x)(x+1)^2' confuses me slightly. Shouldn't the y-intercept be -2?
    y = (2-0)(0+1)(0+1)
    y = (-2)(1)(1)
    y = -2
    What makes the y-intercept positive 2 instead of negative 2?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • duskpin tree style avatar for user meet05
    Posted 5 years ago. Direct link to meet05's post “You should put a practice...”
    You should put a practice for this topic
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user muhammed
    Posted 4 years ago. Direct link to muhammed's post “i cant understand the sec...”
    i cant understand the second question 2) Which of the following could be the graph of y=(2-x)(x+1)^2y=(2−x)(x+1)
    2
    (3 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user ArrowJLC
      Posted 4 years ago. Direct link to ArrowJLC's post “Well you could start by l...”
      Well you could start by looking at the possible zeros. Since the factors are (2-x), (x+1), and (x+1) (because it's squared) then there are two zeros, one at x=2, and the other at x=-1 (because these values make 2-x and x+1 equal to zero).

      Since (x+1) is squared, it has multiplicity 2, which means there's two of them in the factor list. This results in the line of the graph just barely touching zero, rather than crossing it. So you're looking for a graph with zeros at x=-1 and x=2, crossing zero only at x=2.
      Then you determine the end behavior by multiplying all the factors out using algebra, and it has a negative leading coefficient and an odd exponent, which means the end behavior will be x -> (inf) y -> (-inf), and x -> (-inf) y -> (inf).

      Hope this helps!!
      (2 votes)
  • blobby green style avatar for user 23gswansonj
    Posted 4 years ago. Direct link to 23gswansonj's post “How do you find the end b...”
    How do you find the end behavior of your graph by just looking at the equation. how do you determine if it is to be flipped?
    (3 votes)
    Default Khan Academy avatar avatar for user